Let $f_k:(a,b)\rightarrow\mathbb{R}$ differentiable functions by $1\leq k\leq n$. Let $f:(a,b)^n\rightarrow\mathbb{R}$ defined by $$f(x)=\sum_{k=1}^{n}f_k(x_k)$$ Prove that $f$ is differentiable and calculate its derivative at any point of its domain
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asked Jun 28, 2013 at 14:14
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$\begingroup$ Can you write this as a composition of functions that are obviously continuous? Knowing that a function is continuous if and only if it's components are... $\endgroup$– Edvard FagerholmJun 28, 2013 at 14:18
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$\begingroup$ @EdvardFagerholm: No, that's not quite right. You need to observe that $(a,b)^n \overset{\pi_k}{\longrightarrow} (a,b)\overset{f_k}{\longrightarrow}\mathbb R$ is differentiable. Maybe I'm misinterpreting your use of the term "components." $\endgroup$– Ted ShifrinJun 28, 2013 at 14:22
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$\begingroup$ Sorry, I wrote this quickly and used the wrong word. Continuous should be differentiable. $\endgroup$– Edvard FagerholmJun 28, 2013 at 14:30
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1 Answer
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Look at the functions:
$$(x_1,\ldots,x_n)\mapsto \sum_{k=1}^n x_k$$
and
$$(x_1,\ldots,x_n)\mapsto x_k\mapsto f_k(x_k)$$
Use these to build your function through composition. Now apply the chain rule.
